JEE Main 2023 — Binomial Theorem Question with Solution

Any term in the expansion of ${\left(2x+\frac{1}{x^7}+3x^2\right)}^5$ is given by

$\frac{5!}{a!·b!·c!}{\left(2x\right)}^a{\left(x^{-7}\right)}^b{\left(3x^2\right)}^c$

 where $a+b+c=5 ...\left(\mathrm{i}\right)$

$=2^a{\left(3\right)}^c\left(\frac{5!}{a!·b!·c!}\right)x^a{x}^{-7b}{x}^{2c}$

$=2^a{\left(3\right)}^c\left(\frac{5!}{a!·b!·c!}\right)x^{a-7b+2c}$

Now for term to be independent of $x$, we know

$a-7b+2c=0 ...\left(\mathrm{ii}\right)$

Solving $\left(\mathrm{i}\right) \& \left(\mathrm{ii}\right)$, we get

$b=\frac{c+5}{8}$

Also, $a, b, c\in \left\{1,2,3,4,5\right\}$

So, $a=1, b=1, c=3$

Therefore, the term independent of $x$ will be 

$2^1{\left(3\right)}^3·\frac{5!}{1!·1!·3!}=1080$